- #1
lioric
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Plz could you check this working for me??
Two blocks, A and B (with mass 50 kg and 100 kg, respectively), are connected by a string, as
shown in figure.
The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between
block A and the incline is μk = 0.25. Determine the change in the kinetic energy of block A as
it moves from (C) to (D), a distance of 20 m up the incline (and block B drops down a
distance of 20 m) if the system starts from rest.
Block A
T-μR+mg sinθ= ma
Block B
mg-T=ma
acceleration in both blocks is the same since they are connected directly with no elastic strings
Block A
T=ma+μR-mg sinθ
Block B
T=mg-ma
Substitute for T
mg-ma=ma+μR-mg sinθ
(100 x 9.8) - (100a) = (50a) + (0.25 x 50 x 9.8 x cos 37) - (50 x 9.8 x sin 37)
980-100a-50a=97.83-294.89
-100a-50a=97.83-294.89-980
-150a=-1177.057
a=7.8
is this correct?
Homework Statement
Two blocks, A and B (with mass 50 kg and 100 kg, respectively), are connected by a string, as
shown in figure.
The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between
block A and the incline is μk = 0.25. Determine the change in the kinetic energy of block A as
it moves from (C) to (D), a distance of 20 m up the incline (and block B drops down a
distance of 20 m) if the system starts from rest.
Homework Equations
Block A
T-μR+mg sinθ= ma
Block B
mg-T=ma
The Attempt at a Solution
acceleration in both blocks is the same since they are connected directly with no elastic strings
Block A
T=ma+μR-mg sinθ
Block B
T=mg-ma
Substitute for T
mg-ma=ma+μR-mg sinθ
(100 x 9.8) - (100a) = (50a) + (0.25 x 50 x 9.8 x cos 37) - (50 x 9.8 x sin 37)
980-100a-50a=97.83-294.89
-100a-50a=97.83-294.89-980
-150a=-1177.057
a=7.8
is this correct?